Integrand size = 41, antiderivative size = 153 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {\left (b^2 (2 A+C)-2 a (b B-a C)\right ) \text {arctanh}(\sin (c+d x))}{2 b^3 d}-\frac {2 a \left (A b^2-a (b B-a C)\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^3 \sqrt {a+b} d}+\frac {(b B-a C) \tan (c+d x)}{b^2 d}+\frac {C \sec (c+d x) \tan (c+d x)}{2 b d} \] Output:
1/2*(b^2*(2*A+C)-2*a*(B*b-C*a))*arctanh(sin(d*x+c))/b^3/d-2*a*(A*b^2-a*(B* b-C*a))*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(1/2)/b^ 3/(a+b)^(1/2)/d+(B*b-C*a)*tan(d*x+c)/b^2/d+1/2*C*sec(d*x+c)*tan(d*x+c)/b/d
Result contains complex when optimal does not.
Time = 3.01 (sec) , antiderivative size = 472, normalized size of antiderivative = 3.08 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {\cos (c+d x) (b+a \cos (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-2 \left (2 A b^2-2 a b B+2 a^2 C+b^2 C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \left (2 A b^2-2 a b B+2 a^2 C+b^2 C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {8 a \left (A b^2+a (-b B+a C)\right ) \arctan \left (\frac {(i \cos (c)+\sin (c)) \left (a \sin (c)+(-b+a \cos (c)) \tan \left (\frac {d x}{2}\right )\right )}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}\right ) (i \cos (c)+\sin (c))}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}+\frac {b^2 C}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 b (b B-a C) \sin \left (\frac {d x}{2}\right )}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {b^2 C}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 b (b B-a C) \sin \left (\frac {d x}{2}\right )}{\left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{2 b^3 d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) (a+b \sec (c+d x))} \] Input:
Integrate[(Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b* Sec[c + d*x]),x]
Output:
(Cos[c + d*x]*(b + a*Cos[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2) *(-2*(2*A*b^2 - 2*a*b*B + 2*a^2*C + b^2*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*(2*A*b^2 - 2*a*b*B + 2*a^2*C + b^2*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (8*a*(A*b^2 + a*(-(b*B) + a*C))*ArcTan[((I*Cos[c] + S in[c])*(a*Sin[c] + (-b + a*Cos[c])*Tan[(d*x)/2]))/(Sqrt[a^2 - b^2]*Sqrt[(C os[c] - I*Sin[c])^2])]*(I*Cos[c] + Sin[c]))/(Sqrt[a^2 - b^2]*Sqrt[(Cos[c] - I*Sin[c])^2]) + (b^2*C)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (4*b*( b*B - a*C)*Sin[(d*x)/2])/((Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - (b^2*C)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (4*b*(b*B - a*C)*Sin[(d*x)/2])/((Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/(2*b^3*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*(a + b*Sec[c + d*x]))
Time = 1.09 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.07, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.293, Rules used = {3042, 4580, 3042, 4570, 3042, 4486, 3042, 4257, 4318, 3042, 3138, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 4580 |
| \(\displaystyle \frac {\int \frac {\sec (c+d x) \left (2 (b B-a C) \sec ^2(c+d x)+b (2 A+C) \sec (c+d x)+a C\right )}{a+b \sec (c+d x)}dx}{2 b}+\frac {C \tan (c+d x) \sec (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (2 (b B-a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+b (2 A+C) \csc \left (c+d x+\frac {\pi }{2}\right )+a C\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b}+\frac {C \tan (c+d x) \sec (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 4570 |
| \(\displaystyle \frac {\frac {\int \frac {\sec (c+d x) \left (a b C+\left (b^2 (2 A+C)-2 a (b B-a C)\right ) \sec (c+d x)\right )}{a+b \sec (c+d x)}dx}{b}+\frac {2 (b B-a C) \tan (c+d x)}{b d}}{2 b}+\frac {C \tan (c+d x) \sec (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a b C+\left (b^2 (2 A+C)-2 a (b B-a C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {2 (b B-a C) \tan (c+d x)}{b d}}{2 b}+\frac {C \tan (c+d x) \sec (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 4486 |
| \(\displaystyle \frac {\frac {\frac {\left (b^2 (2 A+C)-2 a (b B-a C)\right ) \int \sec (c+d x)dx}{b}-\frac {2 a \left (A b^2-a (b B-a C)\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{b}}{b}+\frac {2 (b B-a C) \tan (c+d x)}{b d}}{2 b}+\frac {C \tan (c+d x) \sec (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\left (b^2 (2 A+C)-2 a (b B-a C)\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{b}-\frac {2 a \left (A b^2-a (b B-a C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{b}+\frac {2 (b B-a C) \tan (c+d x)}{b d}}{2 b}+\frac {C \tan (c+d x) \sec (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\frac {\left (b^2 (2 A+C)-2 a (b B-a C)\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a \left (A b^2-a (b B-a C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{b}+\frac {2 (b B-a C) \tan (c+d x)}{b d}}{2 b}+\frac {C \tan (c+d x) \sec (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle \frac {\frac {\frac {\left (b^2 (2 A+C)-2 a (b B-a C)\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{b^2}}{b}+\frac {2 (b B-a C) \tan (c+d x)}{b d}}{2 b}+\frac {C \tan (c+d x) \sec (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\left (b^2 (2 A+C)-2 a (b B-a C)\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{b^2}}{b}+\frac {2 (b B-a C) \tan (c+d x)}{b d}}{2 b}+\frac {C \tan (c+d x) \sec (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {\frac {\left (b^2 (2 A+C)-2 a (b B-a C)\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {4 a \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{b^2 d}}{b}+\frac {2 (b B-a C) \tan (c+d x)}{b d}}{2 b}+\frac {C \tan (c+d x) \sec (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {\frac {\left (b^2 (2 A+C)-2 a (b B-a C)\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {4 a \left (A b^2-a (b B-a C)\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}}{b}+\frac {2 (b B-a C) \tan (c+d x)}{b d}}{2 b}+\frac {C \tan (c+d x) \sec (c+d x)}{2 b d}\) |
Input:
Int[(Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x]),x]
Output:
(C*Sec[c + d*x]*Tan[c + d*x])/(2*b*d) + ((((b^2*(2*A + C) - 2*a*(b*B - a*C ))*ArcTanh[Sin[c + d*x]])/(b*d) - (4*a*(A*b^2 - a*(b*B - a*C))*ArcTanh[(Sq rt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]*d))/b + (2*(b*B - a*C)*Tan[c + d*x])/(b*d))/(2*b)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo l] :> Simp[1/b Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[( e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[B/b Int[Csc[e + f*x], x], x] + Simp[(A*b - a*B)/b Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x ] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]
Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e _.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_S ymbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2) )), x] + Simp[1/(b*(m + 2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[ b*A*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[ (e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x _Symbol] :> Simp[(-C)*Csc[e + f*x]*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[a*C + b*(C*(m + 2) + A*(m + 3))*Csc[e + f*x] - (2*a*C - b*B* (m + 3))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] & & NeQ[a^2 - b^2, 0] && !LtQ[m, -1]
Time = 0.59 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.61
| method | result | size |
| derivativedivides | \(\frac {\frac {C}{2 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 B b -2 C a -C b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-2 A \,b^{2}+2 B a b -2 C \,a^{2}-C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{3}}-\frac {2 a \left (A \,b^{2}-B a b +C \,a^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{3} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {C}{2 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 B b -2 C a -C b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (2 A \,b^{2}-2 B a b +2 C \,a^{2}+C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b^{3}}}{d}\) | \(247\) |
| default | \(\frac {\frac {C}{2 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 B b -2 C a -C b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-2 A \,b^{2}+2 B a b -2 C \,a^{2}-C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{3}}-\frac {2 a \left (A \,b^{2}-B a b +C \,a^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{3} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {C}{2 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 B b -2 C a -C b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (2 A \,b^{2}-2 B a b +2 C \,a^{2}+C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b^{3}}}{d}\) | \(247\) |
| risch | \(-\frac {i \left (C b \,{\mathrm e}^{3 i \left (d x +c \right )}-2 B b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 C a \,{\mathrm e}^{2 i \left (d x +c \right )}-C b \,{\mathrm e}^{i \left (d x +c \right )}-2 B b +2 C a \right )}{b^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d b}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B a}{d \,b^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C \,a^{2}}{d \,b^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d b}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, d b}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B}{\sqrt {a^{2}-b^{2}}\, d \,b^{2}}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, d \,b^{3}}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, d b}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B}{\sqrt {a^{2}-b^{2}}\, d \,b^{2}}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, d \,b^{3}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d b}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B a}{d \,b^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C \,a^{2}}{d \,b^{3}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d b}\) | \(714\) |
Input:
int(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x,method =_RETURNVERBOSE)
Output:
1/d*(1/2*C/b/(tan(1/2*d*x+1/2*c)-1)^2-1/2*(2*B*b-2*C*a-C*b)/b^2/(tan(1/2*d *x+1/2*c)-1)+1/2/b^3*(-2*A*b^2+2*B*a*b-2*C*a^2-C*b^2)*ln(tan(1/2*d*x+1/2*c )-1)-2*a*(A*b^2-B*a*b+C*a^2)/b^3/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2 *d*x+1/2*c)/((a+b)*(a-b))^(1/2))-1/2*C/b/(tan(1/2*d*x+1/2*c)+1)^2-1/2*(2*B *b-2*C*a-C*b)/b^2/(tan(1/2*d*x+1/2*c)+1)+1/2*(2*A*b^2-2*B*a*b+2*C*a^2+C*b^ 2)/b^3*ln(tan(1/2*d*x+1/2*c)+1))
Leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (140) = 280\).
Time = 7.78 (sec) , antiderivative size = 657, normalized size of antiderivative = 4.29 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx =\text {Too large to display} \] Input:
integrate(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="fricas")
Output:
[1/4*(2*(C*a^3 - B*a^2*b + A*a*b^2)*sqrt(a^2 - b^2)*cos(d*x + c)^2*log((2* a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 - 2*sqrt(a^2 - b^2)*(b*cos (d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos (d*x + c) + b^2)) + (2*C*a^4 - 2*B*a^3*b + (2*A - C)*a^2*b^2 + 2*B*a*b^3 - (2*A + C)*b^4)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (2*C*a^4 - 2*B*a^3* b + (2*A - C)*a^2*b^2 + 2*B*a*b^3 - (2*A + C)*b^4)*cos(d*x + c)^2*log(-sin (d*x + c) + 1) + 2*(C*a^2*b^2 - C*b^4 - 2*(C*a^3*b - B*a^2*b^2 - C*a*b^3 + B*b^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^3 - b^5)*d*cos(d*x + c)^2), -1 /4*(4*(C*a^3 - B*a^2*b + A*a*b^2)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2 )*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c)))*cos(d*x + c)^2 - (2*C*a ^4 - 2*B*a^3*b + (2*A - C)*a^2*b^2 + 2*B*a*b^3 - (2*A + C)*b^4)*cos(d*x + c)^2*log(sin(d*x + c) + 1) + (2*C*a^4 - 2*B*a^3*b + (2*A - C)*a^2*b^2 + 2* B*a*b^3 - (2*A + C)*b^4)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) - 2*(C*a^2* b^2 - C*b^4 - 2*(C*a^3*b - B*a^2*b^2 - C*a*b^3 + B*b^4)*cos(d*x + c))*sin( d*x + c))/((a^2*b^3 - b^5)*d*cos(d*x + c)^2)]
\[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \] Input:
integrate(sec(d*x+c)**2*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c)), x)
Output:
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**2/(a + b*s ec(c + d*x)), x)
Exception generated. \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (140) = 280\).
Time = 0.32 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.88 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {\frac {{\left (2 \, C a^{2} - 2 \, B a b + 2 \, A b^{2} + C b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{3}} - \frac {{\left (2 \, C a^{2} - 2 \, B a b + 2 \, A b^{2} + C b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{3}} - \frac {4 \, {\left (C a^{3} - B a^{2} b + A a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} b^{3}} + \frac {2 \, {\left (2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} b^{2}}}{2 \, d} \] Input:
integrate(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="giac")
Output:
1/2*((2*C*a^2 - 2*B*a*b + 2*A*b^2 + C*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^3 - (2*C*a^2 - 2*B*a*b + 2*A*b^2 + C*b^2)*log(abs(tan(1/2*d*x + 1/2* c) - 1))/b^3 - 4*(C*a^3 - B*a^2*b + A*a*b^2)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1 /2*c))/sqrt(-a^2 + b^2)))/(sqrt(-a^2 + b^2)*b^3) + 2*(2*C*a*tan(1/2*d*x + 1/2*c)^3 - 2*B*b*tan(1/2*d*x + 1/2*c)^3 + C*b*tan(1/2*d*x + 1/2*c)^3 - 2*C *a*tan(1/2*d*x + 1/2*c) + 2*B*b*tan(1/2*d*x + 1/2*c) + C*b*tan(1/2*d*x + 1 /2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*b^2))/d
Time = 20.15 (sec) , antiderivative size = 5489, normalized size of antiderivative = 35.88 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\text {Too large to display} \] Input:
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^2*(a + b/cos(c + d*x))),x)
Output:
((tan(c/2 + (d*x)/2)*(2*B*b - 2*C*a + C*b))/b^2 + (tan(c/2 + (d*x)/2)^3*(2 *C*a - 2*B*b + C*b))/b^2)/(d*(tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^ 2 + 1)) + (atan(((((8*tan(c/2 + (d*x)/2)*(4*A^2*b^7 - 8*C^2*a^7 + C^2*b^7 - 12*A^2*a*b^6 - 3*C^2*a*b^6 + 16*C^2*a^6*b + 16*A^2*a^2*b^5 - 8*A^2*a^3*b ^4 + 4*B^2*a^2*b^5 - 12*B^2*a^3*b^4 + 16*B^2*a^4*b^3 - 8*B^2*a^5*b^2 + 7*C ^2*a^2*b^5 - 13*C^2*a^3*b^4 + 16*C^2*a^4*b^3 - 16*C^2*a^5*b^2 + 4*A*C*b^7 - 8*A*B*a*b^6 - 12*A*C*a*b^6 - 4*B*C*a*b^6 + 16*B*C*a^6*b + 24*A*B*a^2*b^5 - 32*A*B*a^3*b^4 + 16*A*B*a^4*b^3 + 20*A*C*a^2*b^5 - 28*A*C*a^3*b^4 + 32* A*C*a^4*b^3 - 16*A*C*a^5*b^2 + 12*B*C*a^2*b^5 - 20*B*C*a^3*b^4 + 28*B*C*a^ 4*b^3 - 32*B*C*a^5*b^2))/b^4 + (((8*(4*A*b^10 + 2*C*b^10 + 4*A*a^2*b^8 + 8 *B*a^2*b^8 - 4*B*a^3*b^7 + 2*C*a^2*b^8 - 6*C*a^3*b^7 + 4*C*a^4*b^6 - 8*A*a *b^9 - 4*B*a*b^9 - 2*C*a*b^9))/b^6 + (8*tan(c/2 + (d*x)/2)*(C*a^2 + b^2*(A + C/2) - B*a*b)*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/b^7)*(C*a^2 + b^2*(A + C/2) - B*a*b))/b^3)*(C*a^2 + b^2*(A + C/2) - B*a*b)*1i)/b^3 + (((8*tan(c /2 + (d*x)/2)*(4*A^2*b^7 - 8*C^2*a^7 + C^2*b^7 - 12*A^2*a*b^6 - 3*C^2*a*b^ 6 + 16*C^2*a^6*b + 16*A^2*a^2*b^5 - 8*A^2*a^3*b^4 + 4*B^2*a^2*b^5 - 12*B^2 *a^3*b^4 + 16*B^2*a^4*b^3 - 8*B^2*a^5*b^2 + 7*C^2*a^2*b^5 - 13*C^2*a^3*b^4 + 16*C^2*a^4*b^3 - 16*C^2*a^5*b^2 + 4*A*C*b^7 - 8*A*B*a*b^6 - 12*A*C*a*b^ 6 - 4*B*C*a*b^6 + 16*B*C*a^6*b + 24*A*B*a^2*b^5 - 32*A*B*a^3*b^4 + 16*A*B* a^4*b^3 + 20*A*C*a^2*b^5 - 28*A*C*a^3*b^4 + 32*A*C*a^4*b^3 - 16*A*C*a^5...
Time = 0.16 (sec) , antiderivative size = 534, normalized size of antiderivative = 3.49 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x)
Output:
( - 4*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/ sqrt( - a**2 + b**2))*sin(c + d*x)**2*a**3*c + 4*sqrt( - a**2 + b**2)*atan ((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*a**3*c + 2*cos(c + d*x)*sin(c + d*x)*a**3*b*c - 2*cos(c + d*x)*sin(c + d*x)*a**2*b* *3 - 2*cos(c + d*x)*sin(c + d*x)*a*b**3*c + 2*cos(c + d*x)*sin(c + d*x)*b* *5 - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**4*c + log(tan((c + d*x )/2) - 1)*sin(c + d*x)**2*a**2*b**2*c + log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**4*c + 2*log(tan((c + d*x)/2) - 1)*a**4*c - log(tan((c + d*x)/2) - 1)*a**2*b**2*c - log(tan((c + d*x)/2) - 1)*b**4*c + 2*log(tan((c + d*x) /2) + 1)*sin(c + d*x)**2*a**4*c - log(tan((c + d*x)/2) + 1)*sin(c + d*x)** 2*a**2*b**2*c - log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b**4*c - 2*log(t an((c + d*x)/2) + 1)*a**4*c + log(tan((c + d*x)/2) + 1)*a**2*b**2*c + log( tan((c + d*x)/2) + 1)*b**4*c - sin(c + d*x)*a**2*b**2*c + sin(c + d*x)*b** 4*c)/(2*b**3*d*(sin(c + d*x)**2*a**2 - sin(c + d*x)**2*b**2 - a**2 + b**2) )